(x^5 + x^2 - 6x + 5)^1/5 - x = 0

First we will move x to the right side:

==> (x^5 + x^2 - 6x + 5 )^1/5 = x

Now raise to the power 5:

==> x^5 + x^2 - 6x + 5 = x^5

Now subtract x^5 from both sides:

==> x^2 - 6x + 5 = 0

Now factor:

==> (x-5)(x-1) = 0

**Then we have two solutions:**

**==> x1= 5**

**==> x2= 1**

From the given equation (x^5+x^2-6x+5)^(1/5)-x=0

We can get (x^5+x^2-6x+5)^(1/5)=x

=>(x^5+x^2-6x+5)=x^5

=>x^5-x^5+x^2-6x+5=0

=>x^2-6x+5=0

=>x^2-5x-x+5=0

=>x(x-5)-1(x-5)=0

=>(x-1)(x-5)=0

or x=1 and x=5

**Therefore x is equal to 1 and 5.**

First, we notice that there are no constraint of existence of the radical, because the exponent of the radical is odd.

We'll solve the equation by moving the term x to the right side, isolating the radical to the left side.

We'll raise to 5 exponent, expressions from both sides:

[( x^5 + x^2 - 6x + 5 )^1/5]^5 = x^5

x^5 + x^2 - 6x + 5 = x^5

We'll eliminate like terms:

x^2 - 6x + 5 = 0

We'll apply quadratic formula:

x1 = [6+sqrt(36-20)]/2

x1 = (6+4)/2

**x1 = 5**

x2 = (6-4)/2

**x2 = 1**

**Since there are no constraints of existence of the radical, both solutions are valid.**

To solve (x^5+x^2-6x+5)^(1/5) - x = 0

Re write the given equation as:

(x^5+x^2-6x+5 )^1/5 = x.

Raise to the 5th power:

x^5-6x+5 = x^5.

Subtract x^5 from both sides.

x^2-6x+5 = 0

x^2-5x-x+5 = 0

x(x-5)-1(x-5) = 0

(x-5)(x-1) = 0.

So x-5 = 0 or x-1 = 0

x= 5 or x= 1.